Math  /  Algebra

QuestionUse logb20.393,logb30.552\log _{b} 2 \approx 0.393, \log _{b} 3 \approx 0.552, and logb50.801\log _{b} 5 \approx 0.801 to approximate the value of the given logarithm to 3 decimal places. Assume that b>0b>0 and b1b \neq 1. logb15\log _{b} 15 \approx

Studdy Solution

STEP 1

1. We are given approximations for logb2\log_b 2, logb3\log_b 3, and logb5\log_b 5.
2. We need to use properties of logarithms to express logb15\log_b 15 in terms of these known values.
3. The base bb is positive and not equal to 1.

STEP 2

1. Express 15 as a product of the given numbers.
2. Use the properties of logarithms to break down logb15\log_b 15.
3. Substitute the given approximations.
4. Calculate the approximate value.

STEP 3

Express 15 as a product of the given numbers:
15=3×5 15 = 3 \times 5

STEP 4

Use the properties of logarithms to express logb15\log_b 15:
logb15=logb(3×5) \log_b 15 = \log_b (3 \times 5)
Using the property logb(xy)=logbx+logby\log_b (xy) = \log_b x + \log_b y, we have:
logb15=logb3+logb5 \log_b 15 = \log_b 3 + \log_b 5

STEP 5

Substitute the given approximations:
logb30.552 \log_b 3 \approx 0.552 logb50.801 \log_b 5 \approx 0.801
Thus:
logb150.552+0.801 \log_b 15 \approx 0.552 + 0.801

STEP 6

Calculate the approximate value:
logb150.552+0.801=1.353 \log_b 15 \approx 0.552 + 0.801 = 1.353
The approximate value of logb15\log_b 15 is 1.353\boxed{1.353}.

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